Integral extensions of fields

Remarks

Suppose $A$ is a field and let $b \in B$ be nonzero. Take a monic relation $b^n + a_{n-1} b^{n-1} + \cdots + a_0 = 0$ of least degree; since $B$ is a domain, $a_0 \ne 0$. Then $b^{-1} = -a_0^{-1}(b^{n-1} + a_{n-1} b^{n-2} + \cdots + a_1)$ lies in $B$. Conversely, suppose $B$ is a field and let $a \in A$ be nonzero. Then $a^{-1} \in B$ is integral over $A$, say $a^{-m} + c_{m-1} a^{-m+1} + \cdots + c_0 = 0$, and multiplying by $a^{m-1}$ gives $a^{-1} = -(c_{m-1} + c_{m-2} a + \cdots + c_0 a^{m-1}) \in A$.

Example: $\Q \subseteq \Q(\sqrt{2})$ is integral and both are fields; $\Z \subseteq \Z[i]$ is integral and neither is a field.

The hypothesis that $B$ is a domain is needed for "$A$ field implies $B$ field": the ring $k[x]/(x^2)$ is integral over the field $k$, but it is not a field.

Going deeper

For a student

This is the step that links maximal ideals across an integral extension: for a prime $\mathfrak{q}$ of $B$, apply it to $A/(\mathfrak{q} \cap A) \subseteq B/\mathfrak{q}$. It gives that $\mathfrak{q}$ is maximal if and only if $\mathfrak{q} \cap A$ is maximal, and it is used in the proofs of lying over, incomparability and Zariski's lemma.